一、连续登录问题

问题：1）、每个用户连续登录最大天数

            2）、连续登录大于三天的用户数

分析：本质都是计算用户连续登录天数

方案一：利用排序窗口

select a.user_id,a.date_rslt,count(1) as cnt
from (select    t.user_id,t.login_time,date_sub(login_time, num) as date_rsltfrom (select user_id,login_time,row_number() over(partition by user_id order by login_time) as numfrom login_log) t) a
group by a.user_id,a.date_rslt

方案二、增量加全量

连续访问天数v_days（最新flag值为1，则v_days累加，否则为0）

历史最大访问天数max_days (从max_days、v_days中取最大值)

select coaleasce(h.user_id,i.user_id) as user_id,if(i.user_id is not null,v_days+1,0) as v_days,greatest(max_days,if(i.user_id is not null,v_days+1,0)) as max_days
from
history_ds h
full join 
log_time i

扩展1：连续登录，中间间隔1天也算

select user_id,group_id,count(login_date) as continuous_login_daysfrom (select user_id,login_date,sum(if(date_diff>1,1,0)) over(partition by user_id order by login_date rows between unboundedpreceding and current row) as group_idfrom (select user_id,login_date,datediff(login_date,last_login_date) as date_difffrom (select user_id,login_date,lag(login_date,1,'1970-01-01') over(partition by user_id order by login_date) as last_login_datefrom test_login)t1)t2
)t3
group by user_id,group_id;

扩展2:断点排序

连续日期的数据对应的值发生变化，重新排序

select  a,b,row_number() over( partition by b,a_diff order by a) as c
from 
(select  a,b,a-num as a_difffrom (select a,b,row_number() over( partition by b order by  a ) as numfrom t1 )tmp1
)tmp2
order by a; 

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